Question 1 of 8: Number of Stages for a Gas Turbine

Question 1 — Number of Stages for a Gas Turbine (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A single-shaft carbon-dioxide turbine expanding from 21 MPa and 490°C to 7 MPa and 341°C at 1 937 kg/s and 3 600 rev/min, with impulse blading of root-radius-to-height ratio 4 and a constant axial velocity of 100 m/s.

Find. The annulus area the flow needs at the turbine inlet, the mean rotor diameter that follows from it, the blade and gas velocities at that diameter, the gross turbine power and the number of impulse stages required to absorb the whole enthalpy drop.

Approach. Get the inlet density from the ideal-gas law using the paper's own $c_p$ and $c_v$, turn the mass flow into an annulus area by continuity, convert that area into a mean diameter using the fixed $h/r$ ratio, then build the first-stage velocity triangle and divide the total enthalpy drop by the work one impulse stage can extract.

1. Fix the gas properties from the specific heats the paper supplies. The specific gas constant and the ratio of specific heats follow directly: $$R = c_p - c_v = 844 - 655 = 189\ \text{J}\,\text{kg}^{-1}\text{K}^{-1},\qquad k = \frac{c_p}{c_v} = \frac{844}{655} = 1.2886$$ Carbon dioxide is a triatomic gas, so $R$ is much smaller and $k$ much lower than the corresponding air values — which is precisely why it is attractive as a closed-cycle working fluid: for a given pressure ratio the temperature swing is smaller and the machine is more compact.
2. Density and flow area at the turbine inlet (part a). At 21 MPa and 763.15 K the gas is dense: $$\rho_1 = \frac{p_1}{R T_1} = \frac{21\times 10^{6}}{189 \times 763.15} = 145.6\ \text{kg}\,\text{m}^{-3}$$ Continuity through the inlet annulus with the axial velocity fixed at 100 m/s then gives $$\boxed{A_1 = \frac{M}{\rho_1 V_A} = \frac{1937}{145.6 \times 100} = 0.1331\ \text{m}^2}$$ That is a startlingly small duct for nearly two tonnes of gas per second, and it is the whole attraction of a supercritical carbon-dioxide cycle: at this density the turbine annulus is about the size of a dinner plate.
3. Turn the area into a mean rotor diameter (part b). With $h = r/4$ the tip radius is $r_t = r + h = 1.25\,r$, so the annulus area depends on the root radius alone: $$A_1 = \pi\left(r_t^{\,2} - r^{2}\right) = \pi r^{2}\left(1.25^{2} - 1\right) = 0.5625\,\pi r^{2}$$ Solving for the root radius and working outwards, $$r = \sqrt{\frac{0.1331}{0.5625\pi}} = 0.2744\ \text{m},\qquad h = \frac{r}{4} = 0.0686\ \text{m},\qquad r_t = 0.3430\ \text{m}$$ The mean radius sits at mid-blade-height, $r_m = r + h/2 = 1.125\,r$, so $$\boxed{D_m = 2 r_m = 2.25\,r = 0.6174\ \text{m}}$$ This is a free audit of everything above: Question 2 hands the rotor diameter at blade mid-height back as given data, and prints 0.617 m.
4. Blade velocity and nozzle-exit gas velocity (part c). The blade speed is taken at the mean diameter, because that is where a single velocity triangle represents the whole blade: $$V_B = \frac{\pi D_m N}{60} = \frac{\pi \times 0.6174 \times 3600}{60} = 116.4\ \text{m}\,\text{s}^{-1}$$ and the design guideline then fixes the gas velocity leaving the fixed blades: $$\boxed{V_B = 116.4\ \text{m}\,\text{s}^{-1},\qquad V_{S1} = 2.33\,V_B = 271.1\ \text{m}\,\text{s}^{-1}}$$ Using the tip diameter here instead of the mean would overstate $V_B$ by 11 per cent and the stage work by rather more, because the work goes as $V_B$ times a whirl change that itself depends on $V_B$.
5. Gross power of the turbine (part d). The steady-flow energy equation for an adiabatic expansion gives the shaft work directly from the temperature drop, with no need for the pressures at all: $$\boxed{P = M c_p (T_1 - T_2) = 1937 \times 844 \times (490 - 341) = 2.436\times 10^{8}\ \text{W} = 243.6\ \text{MW}}$$ This is the turbine's gross output. In the ARC-100 layout of page 12 the compressor is on the same shaft and absorbs a large share of it, so the net electrical output of the plant is very much smaller — the question is careful to ask only for the turbine.
6. Build the first-stage velocity triangle. The axial component of every absolute velocity is $V_A = 100\ \text{m}\,\text{s}^{-1}$, and page 13 measures the nozzle angle $\theta$ from the plane of rotation, so $$\sin\theta = \frac{V_A}{V_{S1}} = \frac{100}{271.1} = 0.3688 \quad\Rightarrow\quad \theta = 21.6^{\circ}$$ The whirl (tangential) component of the gas leaving the fixed blades is therefore $$V_{w1} = V_{S1}\cos\theta = 271.1 \times 0.9295 = 252.0\ \text{m}\,\text{s}^{-1}$$
7. Work extracted by one impulse stage. Pure impulse blading with no friction means the relative velocity leaves the moving blades with the same magnitude it entered, with its tangential component reversed. The whirl therefore changes by twice the relative inlet whirl, and the work per unit mass is $$w = V_B \Delta V_w = 2 V_B \left(V_{w1} - V_B\right) = 2 \times 116.4 \times (252.0 - 116.4) = 31.57\ \text{kJ}\,\text{kg}^{-1}$$
8. Divide the total drop by the stage work (part e). The whole expansion releases $$\Delta h_{\text{total}} = c_p (T_1 - T_2) = 844 \times 149 = 125.76\ \text{kJ}\,\text{kg}^{-1}$$ so the number of stages required is $$n = \frac{\Delta h_{\text{total}}}{w} = \frac{125.76}{31.57} = 3.98 \quad\Rightarrow\quad \boxed{n = 4\ \text{stages}}$$ Landing within half a per cent of a whole number is the sign that the paper's three guidelines — $h = r/4$, $V_A = 100\ \text{m}\,\text{s}^{-1}$ and $V_{S1} = 2.33\,V_B$ — were chosen to be mutually consistent, and Question 2 confirms it by stating four stages as given data.